The equation of hyperbola $H$ is $\dfrac {(x-4)^{2}}{64}-\dfrac {(y+2)^{2}}{36} = 1$. What are the asymptotes?
Answer: We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac {(y+2)^{2}}{36} = - 1 + \dfrac {(x-4)^{2}}{64}$ Multiply both sides of the equation by $36$ $(y+2)^{2} = { - 36 + \dfrac{ (x-4)^{2} \cdot 36 }{64}}$ Take the square root of both sides. $\sqrt{(y+2)^{2}} = \pm \sqrt { - 36 + \dfrac{ (x-4)^{2} \cdot 36 }{64}}$ $ y + 2 = \pm \sqrt { - 36 + \dfrac{ (x-4)^{2} \cdot 36 }{64}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y + 2 \approx \pm \sqrt {\dfrac{ (x-4)^{2} \cdot 36 }{64}}$ $y + 2 \approx \pm \left(\dfrac{6 \cdot (x - 4)}{8}\right)$ Subtract $2$ from both sides and rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{3}{4}(x - 4) -2$